【題目】

Given two words word1 and word2, find theminimum number of steps required to convert word1 to word2. (each operation iscounted as 1 step.)

You have the following 3 operationspermitted on a word:

a) Insert a character

b) Delete a character

c) Replace a character

word1最少經過多少步可以變成word2。word1允許的操作如下：

a)插入一個字元；

b)刪除一個字元；

c)替換一個字元。

比如“abc”變成“aba”只需要用"a"將最後一個字元"c"替換掉就可以。此時distance = 1

如word1[i] == word2[j],只需把word1[0至i-1]變為word2[0至j-1]。此時res[i][j] = res[i-1][j-1]

【思路】

l  替換：

若word1[i]替換為word2[j]是從word1到word2的最小改動，則需先把word1[0至i-1]以最小距離變為word2[0至j-1]，然後在執行替換操作，此時res[i][j] = res[i-1][j-1] + 1

l  插入：

若word1[i]插入某字元為是從word1到word2的最小改動，則需先把word1[0至i-1]以最小距離變為word2[0至j-1]，然後在執行插入操作，此時res[i][j] = res[i-1][j-1] + 1

l  刪除：

若word1[i]刪除某字元是從word1到word2的最小改動，則需先把word1[0至i-1]以最小距離變為word2[0至j-1]，然後在執行插入操作，此時res[i][j] = res[i-1][j-1] + 1

【Python實現】

class Solution:

    #@return an integer

   def minDistance(self, word1, word2):

       m=len(word1)+1

       n=len(word2)+1

       dp = [[0 for i in range(n)] for j in range(m)]#(m+1)*（n+1）二維矩陣

       for i in range(n):

           dp[0][i]=i

       for i in range(m):

           dp[i][0]=i

       for i in range(1,m):

           for j in range(1,n):

                dp[i][j]=min(dp[i-1][j]+1,dp[i][j-1]+1, dp[i-1][j-1]+(0 if word1[i-1]==word2[j-1] else 1))

       return dp[m-1][n-1]

s = Solution()

s.minDistance("bird","ibdr")